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Chapter 6: Q. 20 (page 575)

$\frac{d y}{d x} = 3 - 4 y, y (0) = 2$

Short Answer

Expert verified

The solution is $y (x) = \frac{1}{4} [3 + 5 e^{- 4 x}]$

Step by step solution

01

Step 1. Given information

given differential equation is $\frac{d y}{d x} = 3 - 4 y$

02

Step 2. solve equation

$\int \frac{1}{3 - 4 y} d y = \int d x - \frac{1}{4} \ln (|3 - 4 y|) = x + c 1 \ln (|3 - 4 y|) = - 4 x - 4 c 1 3 - 4 y = A e^{- 4 x} y = \frac{1}{4} (3 - A e^{- 4 x}) 2 = \frac{1}{4} (3 - A) 8 - 3 = - A A = - 5 y (x) = \frac{1}{4} (3 + 5 e^{- 4 x})$

03

Step 3. The solution

The Solution is $y (x) = \frac{1}{4} [3 + 5 e^{- 4 x}]$

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Most popular questions from this chapter

Consider the region between the graph of $f (x) = x - 2$ and the x-axis on [2,5]. For each line of rotation given in Exercises 35– 40, use definite integrals to find the volume of the resulting solid.

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{d y}{d x} = - 2 (1 - 3 y), y (0) = 2$

Sketching a representative disks ,washers and shells : sketch a representative disks , washers , shells for the solid obtained by revolving the regions shown in figure around the given lines .

the line $x = 7$

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52.

\frac{d y}{d x} = 1 - y, y (0) = 4

In the process of solving the differential equation $\frac{d y}{d x} = 1 - y$ by separation of variables, we obtain the equation $- \ln | 1 - y | = x + C$ . After solving for $y$ , this equation becomes $y = 1 - A e^{- x}$ . Given that $y > 1$ , how is A related to $C$ ?

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